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This question already has an answer here:

I could show in calculus where $0<θ<\frac{\pi}{2}$ like this

$$\frac{d(θ)}{dθ}=1<\frac{1}{2}(\cosθ+\sec^2θ)$$

But I curioused about would it could prove by geometric way likewise trigonometric identities. I try to show with the area of circular sector but I failed...

So this is question:

How to show $$\frac{\sinθ +\tanθ}{2}>\theta$$ where $0<\theta<\frac{\pi}{2}$ in geometric way?

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marked as duplicate by Namaste, Blue geometry Aug 23 '18 at 16:39

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Recall that $\theta$ is the length of the arc of radius $\theta$. This may be of help to prove the inequality geometrically on the unit circle $\endgroup$ – b00n heT Aug 23 '18 at 15:19
  • $\begingroup$ Please fix the geomatic/geomatric typos. $\endgroup$ – Yves Daoust Aug 23 '18 at 15:23
  • $\begingroup$ @YvesDaoust I fixed it $\endgroup$ – user366725 Aug 23 '18 at 15:25
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    $\begingroup$ @user366725: I am afraid not. $\endgroup$ – Yves Daoust Aug 23 '18 at 15:26
  • $\begingroup$ If you are interested in more calculus-based proofs, you can check Jack's and Martin's answer here: math.stackexchange.com/questions/2196624/… $\endgroup$ – Botond Aug 23 '18 at 15:33
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Here are some thoughts:

Approach $1a$

enter image description here

The blue region is a unit arc with an angle $\theta$. Let the length of $\overline {IH}$ be $\displaystyle L=\frac{\sin\theta+\tan\theta}2$.

The area of $\triangle AIH$ is given by $$S=\frac12 L\left(\frac L{\tan\theta}\right)=\frac{L^2}{2\tan\theta}\tag{1}$$

Notice the area of $\triangle ADG> \text{arc } ACG\implies\tan\theta>\theta\tag 2$

By $(1), (2)$

$$S<\frac{L^2}{2\theta}$$

Approach 1b

Notice arc $CJG$ has the length $\theta$. Then to prove $\theta<L$ is equivalent to prove the length of arc $CJG$ is shorter than the length of line $\overline {IH}$.

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  • $\begingroup$ To be honest, I don't think my attempt deserves to be an accepted answer. It is incomplete, plus it has a better and complete solution in the previous question. $\endgroup$ – Mythomorphic Aug 26 '18 at 16:07

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